To answer the first part of the question, you need to find a table containing values for normal distribution. Usually, this will be a table of z-scores, which are calculated in the following way:

`z = (x-mu)/sigma`

Here, `x` is the sample value, `mu` is the global mean, and `sigma`...

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To answer the first part of the question, you need to find a table containing values for normal distribution. Usually, this will be a table of z-scores, which are calculated in the following way:

`z = (x-mu)/sigma`

Here, `x` is the sample value, `mu` is the global mean, and `sigma` is the global standard deviation. Let's calculate the z-score for a score of 25.4:

`z = (25.4-21)/3.214 = 1.37`

So, 25.4 is 1.37 standard deviations above the mean. Using the z-table found in the link below, we can now say that 0.913 of the population will be below this score. Therefore, the probability that a student has a score less than 25.4 is 0.913.

Using symmetry of the normal distribution, we can say that if the probability of a student having a score greater than b has the same probability as a student having a score less than 25.4, then the z-score will be reflected across zero. In other words, the z-score for b will be the negative of the z-score for 25.4. this gives us the following z-score:

`z_b = -1.37`

To find b with the z-score, we simply substitute all of our known variables into the z-score formula and solve for the sample value:

`z_b = (b-mu)/sigma`

`-1.37 = (b-21)/3.214`

Multiply both sides by 3.214:

`-4.40318 = b-21`

Add both sides by 21:

`16.60 = b`

You might notice that we were simply taking the difference between 25.4 and the mean, then we subtracted that from the mean. This method would work fine in this situation, but it may not be universal for all questions using these variables.

I hope this helps!